MASSACHUSETTS MATHEMATICS LEAGUE CONTEST 5 FEBRUARY 2013 ROUND 1 ALGEBRA 2: ALGEBRAIC FUNCTIONS ANSWERS

Size: px

Start display at page:

Download "MASSACHUSETTS MATHEMATICS LEAGUE CONTEST 5 FEBRUARY 2013 ROUND 1 ALGEBRA 2: ALGEBRAIC FUNCTIONS ANSWERS"

Roxanne Bryan
5 years ago
Views:

1 CONTEST 5 FEBRUARY 03 ROUND ALGEBRA : ALGEBRAIC FUNCTIONS ANSWERS A) B) (,, ) C) A) Let f( x) 3x f 5 + f 3. =. Compute: ( ) 8 B) Given: f( x) = 3x gx ( ) = ( x )( x+ 3) + A, where A< 0 ( ) has For several integer values of A, the composite function h( x) = g f( x) = g f ( x) two distinct rational zeros, r and r, where r < r. For the largest possible value of A, compute the ordered triple ( Ar,, r ) C) f( x ) is a th degree polynomial with a leading coefficient of. Curiously, f() = f() = f(3) = f() = 6. Determine the sum of the zeros of this function.

2 CONTEST 5 - FEBRUARY 03 ROUND ARITHMETIC / NUMBER THEORY ANSWERS A) B) C) A) How many two-digit primes leave a remainder of when divided by? Assume prime refers to positive integers only. B) The positive integers N and (N + ) have 3 and 6 positive factors respectively. Compute the smallest possible value of N. C) Find the smallest prime factor of

3 CONTEST 5 - FEBRUARY 03 ROUND 3 TRIG: IDENTITIES AND/OR INVERSE FUNCTIONS ANSWERS A) B) C) y = 5 A) Given: Arc sin = tan( ) 7 Arc k Determine the value of k in terms of a simplified radical. If necessary, rationalize the denominator. B) Let A = Arc tan ( 3). Compute sin(π + A). x = 9cos θ C) Given:, where 0 θ < π y = 9sin θ Clearly, 0 x 9 and 0 y 9 Express y strictly in terms of x, where 0 x 9.

4 CONTEST 5 - FEBRUARY 03 ROUND ALG : WORD PROBLEMS ANSWERS A) B) C) A) A square of side N is cut from one corner of a 3 x 5 index card, leaving 85% of the index card (by area). Compute N. B) A jar contains 50 jelly beans, either red (R) or green (G). The initial ratio is R : G = 5 : 3. The red jelly beans are more tasty and disappear twice as fast as the green. How many jelly beans are left in the jar when R : G = 3 : 5? C) A simple majority wins the election for senior class president. Boris and Miles were the final two candidates for senior class president. Boris won the election; however, if exactly 8% of those who voted for Boris switched their vote to Miles and those who voted for Miles originally voted for him again, Miles would have won the election by one vote. More than 00 votes were cast in the election. Compute the minimum possible number of votes that were cast.

5 CONTEST 5 - FEBRUARY 03 ROUND 5 PLANE GEOMETRY: CIRCLES ANSWERS A) B) C) sq. inches E A) Given: ST is a diameter of circle O. m ETS = 53, m SEA = m TEA Compute the degree-measure of minor arc ETA. S. O T A B) In circle O, chords AB and CD intersect at point E. Ray PT is tangent to circle O at point T. Ray PQ pass through point O. CE = 6, DE =, AE = 3, PQ =.5, BE = PD Compute the radius of circle O. O P B Q D E C T A C) Nine circles are inscribed inside a square whose area is 6 square inches. Compute the area of the shaded region.

6 CONTEST 5 - FEBRUARY 03 ROUND 6 ALG : SEQUENCES AND SERIES ANSWERS A) B) : C) A) Let S denote the sum of the 0 th and 3 th terms of the geometric progression 3 3,, 3, 3 i,.... Compute S. i B) 3, y, x are the first three terms in a geometric progression (GP) and y 0. 3, y+, x+ are the first three terms in an arithmetic progression (AP). Compute the ratio of the th term of the AP to the 5 th term of the GP. C) We know that the arithmetic mean is always greater than or equal to the harmonic mean. Suppose for some positive value(s) of A, the arithmetic mean and harmonic mean of and A differ by 0.. Compute all possible positive values of the geometric mean of and A. FYI: The harmonic mean of x and y is defined to be the reciprocal of the arithmetic mean of x and y.

7 CONTEST 5 - FEBRUARY 03 ROUND 7 TEAM QUESTIONS ANSWERS A) D) B) E) (, ) C) F) 3 Ax + Bx 6x + 3 A) Given: f( x) =, f(5) = 7, f( ) = x It has a linear asymptote y = mx + b as x ±. Compute the ordered pair (m, b). B) Let N be a -digit integer consisting exclusively of prime base 0 digits, but not necessarily distinct. How many of these integers are divisible by? cos 5 Arc Arccos x Arccos + = 3 0. C) Compute x such that ( ) D) Suppose a sheet of first-class FOREVER stamps costs $6.7 in 0. Suppose that due to a increase in the rate, a sheet of FOREVER stamps with 8 more stamps cost $.00 in 0. No sheet ever contains more than 50 stamps. If a FOREVER stamp costs 6 in 03, how much more will a FOREVER stamp cost in 0? E) PQRS is a square, PQ = 6. QMS is an arc of a circle with center at R and radius RQ. Circle T is tangent to sides of the square and to the arc at point M. The ratio of the area of circle T to the area of the segment on SQ (i.e. the shaded region) may be expressed as Aπ, where B is an integer. Compute the ordered pair (A, B). π B F) Suppose a sequence is defined by the recursive relation an a n + =. If the first five terms are all positive integers, compute the minimum sum of these five terms.

8 Round Alg : Algebraic Functions CONTEST 5 FEBRUARY 03 ANSWERS A) 5 B) Round Arithmetic/ Number Theory 5 6,, 3 3 C) 0 A) 0 B) 9 C) 7 Round 3 Trig Identities and/or Inverse Functions A) 5 6 B) 3 7 C) y = 3 x or 9 x + x Round Alg : Word Problems Round 5 Geometry: Circles A).5 B) 7 C) 39 A) 6 B) 5.5 C) 7 8π or 8( π ) Round 6 Alg : Sequences and Series A) 8i B) 9 : C) 6, 5 5 Team Round A) (5, ) D) ( 7, ) B) 3 E) ( 68 8, ) or ( ) C) F) 57

9 Round A) CONTEST 5 FEBRUARY 03 SOLUTION KEY f 8 8 = 3 = =. 3 3 Let a f ( 5) Thus, f ( 5) =. Then: f( a ) = 5 3a = 5 3a = 5 a = 3. + f =3 + = 5. 3 f( x) = 3x g f( x) = gx ( ) = ( x )( x+ 3) + A, where A< 0 3x 3x+ + A= 9x 9 x+ ( A ) 8 B) ( )( ) To find the zeros, we use the quadratic formula to solve 9x 9 x+ ( A ) = 0 : 9 ± 8 36( A ) 9 ± 5 36A x = = 8 8 Examining the discriminant, we want the largest possible negative integer for which 5 36A is a perfect square. For A =, 5, we get non-perfect squares 6, 97, 333, 369, and 05; but A = 6 gives us = 9 ± 30 5 Therefore, x =,, Since r < r, we have ( Ar,, r) = 6,, 3 3 The question alluded to several other values for which the composite function had rational zeros. Some other values of A are:,, 36, 50, 66, Do you see a pattern? Investigate. Can you prove a conjecture? 3 f ( x) = x + ax + bx + cx + d C) f() = f() = f(3) = f() = 6 We do not have to determine the actual zeros, since the coefficient a is the opposite of the sum of the zeros a+ 7b+ c= a+ b= 0 Subtracting, a+ 5b+ c= 0 a= a+ b= a+ 3b+ c= 0 Thus, the sum of the roots is 0. FYI: Backtracking with a = 60 in the remaining equations, 3 f( x) = x 0x + 35x 50x+ 30. Alternative: Let f( x) = ( x )( x )( x 3)( x ) + 6 and the given requirements are satisfied and the sum of the zeros is the opposite of the coefficient of x 3, i.e. +0.

10 CONTEST 5 FEBRUARY 03 SOLUTION KEY Round A) Positive integers leaving a remainder of when divided by are of the form N = k +, for k = 3,, 5, Substituting (k, N) = (3, 3), (, 7), (7, 9), (9, 37), (0, ), (, 53), (5, 6), (8, 73), (, 89) and (, 97) 0 two-digit primes B) The only numbers with an odd number of divisors are perfect squares. = (, 5) fails, since 5 is prime 3 = 9 (9, 0) fails, since 0 has divisors 5 = 5 (5, 6) fails, since 6 has divisors 7 = 9, 50 = 5 6 factors Thus, N = C) ( ) + = + = 5 = 0 5 = 0(3) = 3768 We require the smallest prime factor of Obviously, 3 and 5 fail, but 3767 = 7(68). 68 may not be prime, but it does not have a smaller factor than 7. FYI: In fact, 68 is composite, since (3)(5) = = and the same result Also, you may have noticed that ( )( ) follows. Verify that A 5 ( A ) ( A A 3 A A ) = and substitute 3 A =. [In the world of computers, it s very useful to remember (memorize) the fact that K (kilobyte) = 0 = 0 bytes. K means thousand and 0 is the closest power of.] megabyte (Mb) = 0 (,08,576 bytes). M means million and 0 is the closest power of. gigabyte (Gb) = 30 (,073,7,8 bytes). G means billion and 30 is the closest power of. terabyte (Tb) = 0 (,099,5,67,776 bytes). T means trillion and 0 is the closest power of.

11 CONTEST 5 FEBRUARY 03 SOLUTION KEY Round 3 5 A) Let θ = Arcsin 7. This is equivalent to 5 sinθ = and 7 π 0 < θ < (quadrant ), since the argument was positive. AC = 9 5 = Thus, tan θ = 6 = or, equivalently, 5 6 θ = Arc tan and k = A θ B 5 C π π B) The range of y = Arc tan( x) is,. Since A is in quadrant ( tan A = 3), π + A must be in quadrant. The sine in quadrant is positive A 7. P (, - 3) x = 9cos θ C) Since, it is clear that both x and y must be y = 9sin θ nonnegative. For real numbers x and y, x = x and y = y denote nonnegative numbers. Taking the square root and summing, we have x ( ) + y = + = + = = 3cos θ 3sin θ 3 cos θ sin θ 3 3 y = 3 x y = 3 x = 9 x + x What does the graph of this function look like? Compare your graph with the graph is at the end of this solution key. Carefully consider the accompanying commentary. The tail to the right of point A is extraneous for our pair of parametric equations, since the value of x could not be greater than 9. x = 9 cos θ, where 0 θ < π and y = 9sin θ are equivalent only over 0 x 9. For a real-valued function the only restriction on the domain is 0 y = 3 x y = 3 x, x and the graph would continue to the right.

12 CONTEST 5 FEBRUARY 03 SOLUTION KEY Round A) If 85% is left, then 5% was removed. Thus, 5 N = 0.5(5) = N = B) Let (R, G) = (5x, 3x) initially. Then: 8x = 50 x = 63 and there are 35 red jelly beans and 89 green jelly beans at the outset. We require that 35 x 3 = 575 0x= 567 3x 7x= 008 x= 89 x 5 Thus, there are = 7 red jelly beans and 89 = 5 green jelly beans, for a total of 7 jelly beans. C) Let B and M denote votes for Boris and Miles respectively. M, B = (,5), (3,50),(6,75) 39 votes cast. M must be divisible by ( ) Check: This is the first total that exceeds 00. 8% of 75 votes is 6 votes. If Boris loses 6 votes then, instead of a 75 to 6 victory for Boris, the results become a 70 to 69 victory for Miles, a one vote differential as required.

13 Round 5 CONTEST 5 FEBRUARY 03 SOLUTION KEY A) Since SET is inscribed in a semi-circle, it s a right angle. Thus, m SEA = m TEA = 5. As intercepted arcs of SEA and ETS, msa = 90, mse = 06 met = 7 ETA = =6. S 06. O E T A B) Let R denote the second point of intersection of ray PQ and circle O and recall that we were given that PD = BE. Applying the product chord theorem, AE BE = CE DE 3AE = 6 BE = PD = Applying the tangent-secant rule, PD PC = PT PT = ( + 7) = 8 and PQ PR = 8.5(.5 + r) = 8 r+.5 = r =5.5 P B Q D E O C C) Let D denote the diameter of one of the circles. Then: 3D = 6 6R = 9 R = 3 T A From the diagram at the right, we see that the star-shaped region between the circles has an area equal to a square minus a full circle, i.e. ( 3 ) 3 π 9π = 8 Multiplying by, the required area is 7 8π or ( ) 8 π.

14 CONTEST 5 FEBRUARY 03 SOLUTION KEY Round 6 A) The second term of the geometric progression can be written as 3i. The common ratio r for the geometric progression is i = i. Therefore, the progression is 3, 3, i 3,3,3,... i consists of a repetition of four terms. If n is more than a multiple of, i.e. n =, 5, 9,, t n = 3. 3 Since 0 is more than a multiple of, t 0 = = 3i. i Since 3 is more than a multiple of, t 3 = 3 ( i ) = 9 8i+ 9 = 8i (Also accept 0 8i. ) Do not accept 8 i. B) () y = 3x () x + y = y + 3 x = y Thus, y = 6 y yy ( 6) = 0. Since y 0, (x, y) = (, 6) in the arithmetic progression: t = 3 + ( 5 ) = 58 in the geometric progression: ( ) The required ratio is 9: t 5 = 3 = 8 C) A+ A AM = HM + 0. = + Multiplying through by 0(A + ), we have A + 0 5( A+ ) = 0 A+ ( A+ ) 5A A+ 8 = (5A 6)( A 3) = 0 6 Therefore, A = 3,. 5 A = 3 GM = 6 A =. GM = 6 35 = = (or 5 5 )

15 CONTEST 5 - FEBRUARY 03 SOLUTION KEY Team Round A+ B+ 9 f( ) = = A B= 3 5A+ 5B 7 A) f(5) = 7 = 7 5A+ 5B= 99(7) + 7 = 7(99 + ) = A+ B= Thus, (A, B) = (0, 8). By long division, x + 8 x 6 x x + = x + + x x. As x, the value of the fractional third term approaches zero, since the degree of the denominator is more than the degree of the numerator implying the denominator grows much faster than the numerator. Therefore, the functional values become closer and closer to y = 5x + and (m, b) = (5, ). Is the graph of the function above or below the line as x? For other values of x? Think about it and then look at the graph at the end of this solution key. B) The test: A -digit integer a b c d is divisible by if and only if (a + c) (b + d) is divisible by, that is, equal to either 0 or. Sum digits in even positions, sum digits in odd positions and then subtract. The prime digits are, 3, 5 and 7. Case : (all digits the same) All possibilities have (a + c) (b + d) = 0 and are divisible by. Case : (3 digits same) - 3 = digit selections x arrangements 8 possible N-values For any integer of this form, for example, x x x y, (a + c) (b + d) = x y. The minimum and maximum differences are and 5 respectively. None of these N-values are divisible by. Case 3: ( digits same, different) - 6 digit selections x 7 possible N-values For 35, 37, 57 min =,, (sum of largest and smallest minus sum of other two) max =, 6, 8 (sum largest two sum smallest two) None For 335, 337, 3357 min =,, max = 3, 5, 6 None For 553, 557, 5537 min =,,0 max = 5, 5, (7535, 3575, 5357, 5753) For 773, 775, 7735 min =, 3, max = 9, 7, 6 None Case : ( pairs of digits the same) - 6 digit selections x 6 36 possible N-values Consider one of the 6 selections, e.g., 3 3. The required difference is either 6 = or 5 5 = 0. There are arrangements giving a difference of 0, namely 3 3, 3 3, 3 3, 3 3. This is true for each of the 6 selections. Thus, there are possible N-values. Case 5: (all digits different) -! = possibilities Using all distinct digits, 3, 5 and 7, the minimum difference 9 8 = and the maximum difference is 5 = 7. Therefore, none of these are possible N-values. Thus, the total is = 3.

16 CONTEST 5 - FEBRUARY 03 SOLUTION KEY Team Round - continued C) Rewrite 5 Arccos Arccos( x) Arccos + = 3 0 as 5 Arccos( x) = Arccos Arccos 0 3 and let A = Arccos 0 and B = 5 Arccos. Then, taking the cosine of both sides, 3 x = cos(a B) = cos Acos B+ sin Asin B= + = = A B D) Suppose in 0 there are n stamps on the sheet costing c cents each. In 0, the sheet consists of (n + 8) stamps costing (c + ) cents each. nc = 67 Then: n + 8c = 00 3 nc = = 96 or n = c ( n+ 8)( c+ ) = 00 ( c) c= 67 c c+ 67 = ( c 6c+ 336) = ( c 6)( c 56) = 0 c= 6,56 Thus, in 0, there were stamps on the sheet, costing 56 each. ( stamps at 6 is rejected, since > 50.) Thus, the cost of a FOREVER stamp in 0 is 60, more than in 03.

17 Team Round - continued CONTEST 5 - FEBRUARY 03 SOLUTION KEY E) The shaded region plus the region bounded by ΔSQR is a ( ) π 6 quarter circle with area = 9π. Thus, the area of the shaded region is 9π 66 = 9( π ). Draw diagonal PR, intersecting SQ at point O. PR = 6 PO = 3. Since points T and M lie on PR, PT + TM + MO = 3 (***). If r denotes the radius of circle T, PT = r (since P, T and the points of tangency on the square form a small square with side r). Therefore, (***) is equivalent to: ( ) ( ) 6( ) 6( ) r + r = 3 ( MO = RM RO) r + = ( ) ( ) r = = 6 = Finally, the required ratio is ( ) (A, B) = ( 68 8, ) or ( ) ( ) ( ) ( ) π r π 36 3 π 68 8 = = 9 π 9 π π ( 7, ) ( ) is not an acceptable as the computed value of A. and

18 CONTEST 5 - FEBRUARY 03 SOLUTION KEY Team Round - continued an F) an an+ = an+ = Using this form of the recursive definition, we can find expressions for the first five terms, all in terms of a. a a = a a a 3 a3 = = = a 7 a = 8 a 5 a5 = 6 All of these expressions must generate integers. The smallest value of a for which a5 is an integer is 3, producing a 5 = Substituting back up the chain, ( ) a, a, a, a, a = (, 3, 7, 5, 3) and the sum is

19 FYI: The graph of 3C) connects points A and B. The tail to the right of point A is extraneous for our pair of parametric equations, since the value of x could not be greater than 9. x = 9cos θ, where 0 θ < π and y = 9sin θ equivalent only over 0 x 9 y = 3 x are For a real-valued function y = 3 x, the only restriction on the domain is x 0 and the graph would continue to the right. Here s the graph of Team A) The graph always lies above y = 5x + for x< or x> and below for x-values in-between these two critical values.. As x + (to the right) or x (to the left), the graph is almost indistinguishable from the straight line. In fact, this is true for x < 3 and for x > +. x=± and y = 5 x+ are called asymptotes, lines which the function gets arbitrarily close to, but never actually makes contact! Over < x <, the graph reaches a maximum value over the interval 0, and opens downward. Using calculus or a graphing calculator, over this interval, the maximum value of approximately.57 occurs at approximately x =

* NO CALCULATORS ON THIS ROUND *

CONTEST 5 FEBRUARY 0 ROUND ALGEBRA : ALGEBRAIC FUNCTIONS ANSWERS A) B) : C) ***** NO CALCULATORS ON THIS ROUND ***** A) Let 4x+ 5 for x> f( x) = 0 for < x 3x 4 for 0 x f f() + f f( ) + f(0) Compute: (